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Ta có \(A\left(x\right)=\dfrac{1}{3}x+1=0\Leftrightarrow x=-1:\dfrac{1}{3}=-3\)
\(B\left(x\right)=-\dfrac{3}{4}x+\dfrac{1}{3}\Leftrightarrow x=-\dfrac{1}{3}\left(-\dfrac{3}{4}\right)=4\)
\(C=\left(2x-4\right)\left(x+1\right)=0\Leftrightarrow x=2;x=-1\)
\(D\left(x\right)-4x\left(x-2\right)=0\Leftrightarrow x=0;x=2\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a,\Leftrightarrow-\dfrac{1}{2}x=\dfrac{1}{4}\Leftrightarrow x=-\dfrac{1}{2}\\ b,\Leftrightarrow\dfrac{1}{6}:x=\dfrac{1}{2}+\dfrac{1}{3}=\dfrac{5}{6}\Leftrightarrow x=\dfrac{1}{6}:\dfrac{5}{6}=\dfrac{1}{5}\\ c,\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{5}=3\\x+\dfrac{1}{5}=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{14}{5}\\x=-\dfrac{16}{5}\end{matrix}\right.\)
\(d,\Leftrightarrow\left(x+\dfrac{1}{2}\right)^2=\dfrac{22}{9}-\dfrac{7}{3}=\dfrac{1}{9}\\ \Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{2}=\dfrac{1}{3}\\x+\dfrac{1}{2}=-\dfrac{1}{3}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{6}\\x=-\dfrac{5}{6}\end{matrix}\right.\\ e,\Leftrightarrow2\left|x\right|=2-\dfrac{1}{2}=\dfrac{3}{2}\\ \Leftrightarrow\left[{}\begin{matrix}2x=\dfrac{3}{2}\\2x=-\dfrac{3}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{4}\\x=-\dfrac{3}{4}\end{matrix}\right.\)
\(f,\Leftrightarrow\left|x+\dfrac{1}{2}\right|=1+\dfrac{1}{6}=\dfrac{7}{6}\\ \Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{2}=\dfrac{7}{6}\\x+\dfrac{1}{2}=-\dfrac{7}{6}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=-\dfrac{5}{3}\end{matrix}\right.\)
e: ta có: \(2\left|x\right|+\dfrac{1}{2}=2\)
\(\Leftrightarrow2\left|x\right|=\dfrac{3}{2}\)
\(\Leftrightarrow\left|x\right|=\dfrac{3}{4}\)
hay \(x\in\left\{\dfrac{3}{4};-\dfrac{3}{4}\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a: \(5^{\left(x-2\right)\left(x+3\right)}=1\)
=>\(\left(x-2\right)\left(x+3\right)=0\)
=>\(\left[{}\begin{matrix}x-2=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-3\end{matrix}\right.\)
c: \(\left|x^2+2x\right|+\left|y^2-9\right|=0\)
mà \(\left\{{}\begin{matrix}\left|x^2+2x\right|>=0\forall x\\\left|y^2-9\right|>=0\forall y\end{matrix}\right.\)
nên \(\left\{{}\begin{matrix}x^2+2x=0\\y^2-9=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\left(x+2\right)=0\\\left(y-3\right)\left(y+3\right)=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x\in\left\{0;-2\right\}\\y\in\left\{3;-3\right\}\end{matrix}\right.\)
d: \(2^x+2^{x+1}+2^{x+2}+2^{x+3}=120\)
=>\(2^x\left(1+2+2^2+2^3\right)=120\)
=>\(2^x\cdot15=120\)
=>\(2^x=8\)
=>x=3
e: \(\left(x-7\right)^{x+1}-\left(x-7\right)^{x+11}=0\)
=>\(\left(x-7\right)^{x+11}-\left(x-7\right)^{x+1}=0\)
=>\(\left(x-7\right)^{x+1}\left[\left(x-7\right)^{10}-1\right]=0\)
=>\(\left[{}\begin{matrix}x-7=0\\x-7=1\\x-7=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=7\\x=8\\x=6\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Em muốn hỏi bài nào vậy? Quá nhiều bài thầy cô và các bạn không thể trả lời được hết em ạ
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) 1/4(x-3)+2=1/5
1/4.(x-3) = 1/5-2
1/4.(x-3) = -9/5
x-3 = (-9/5):1/4
x-3 = -36/5
x = -36/5+3
x= -21/5
![](https://rs.olm.vn/images/avt/0.png?1311)
a.\(x=0;y=-1\)
\(\Rightarrow2.0-\dfrac{-1\left(0^2-2\right)}{0.-1-1}=0-\dfrac{2}{-1}=2\)
b.\(x=2\)
\(\Rightarrow4.2^2-3\left|2\right|-2=16-6-2=8\)
\(x=-3\)
\(\Rightarrow4.\left(-3\right)^2-3\left|-3\right|-2=36-9-2=25\)
c.\(x=-\dfrac{1}{5};y=-\dfrac{3}{7}\)
\(\Rightarrow5.\left(-\dfrac{1}{5}\right)^2-7.\left(-\dfrac{3}{7}\right)+6=5.\dfrac{1}{25}+3+6=\dfrac{1}{5}+3+6=\dfrac{46}{5}\)
thay x=2 và biểu thức A ta đc
\(A=4.2^2-3.\left|2\right|-2=4.4-6-2=16-6-2=8\)
thay x=-3 biểu thức A ta đc
\(A=4.\left(-3\right)^2-3.\left|-3\right|-2=4.9-9-2=36-9-2=25\)
thay x=-1/5 ; y=-3/7 biểu thức B ta đc
\(B=5.\left(-\dfrac{1}{5}\right)^2-7.\left(-\dfrac{3}{7}\right)+6\)
\(B=5\cdot\dfrac{1}{25}+3+6\)
\(B=\dfrac{1}{5}+3+6=\dfrac{46}{5}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a: \(\Leftrightarrow\left|x-1\right|=3-2x\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(2x-3-x+1\right)\left(2x+3+x-1\right)=0\\x< =\dfrac{3}{2}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left(x-2\right)\left(3x+2\right)=0\\x< =\dfrac{3}{2}\end{matrix}\right.\)
=>x=-2/3
b: Trường hợp 1: x<-3
Pt sẽ là:
\(-x-1-x-3=10-4x\)
=>-2x-4=10-4x
=>2x=14
hay x=7(loại)
Trường hợp 2: -3<=x<-1
Pt sẽ là \(x+3-x-1=10-4x\)
=>10-4x=2
=>4x=8
hay x=2(loại)
Trường hợp 3: x>=-1
Pt sẽ là x+1+x+3=10-4x
=>2x+4=10-4x
=>6x=6
hay x=1(nhận)